Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{3x + 18}{-x^2 + 2x + 48} \div \dfrac{-x - 1}{x^2 - 4x - 5} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{3x + 18}{-x^2 + 2x + 48} \times \dfrac{x^2 - 4x - 5}{-x - 1} $ First factor out any common factors. $q = \dfrac{3(x + 6)}{-(x^2 - 2x - 48)} \times \dfrac{x^2 - 4x - 5}{-(x + 1)} $ Then factor the quadratic expressions. $q = \dfrac {3(x + 6)} {-(x + 6)(x - 8)} \times \dfrac {(x + 1)(x - 5)} {-(x + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {3(x + 6) \times (x + 1)(x - 5) } { -(x + 6)(x - 8) \times -(x + 1)} $ $q = \dfrac {3(x + 1)(x - 5)(x + 6)} {(x + 6)(x - 8)(x + 1)} $ Notice that $(x + 6)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {3(x + 1)(x - 5)\cancel{(x + 6)}} {\cancel{(x + 6)}(x - 8)(x + 1)} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $q = \dfrac {3\cancel{(x + 1)}(x - 5)\cancel{(x + 6)}} {\cancel{(x + 6)}(x - 8)\cancel{(x + 1)}} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $q = \dfrac {3(x - 5)} {x - 8} $ $ q = \dfrac{3(x - 5)}{x - 8}; x \neq -6; x \neq -1 $